On Tue, Dec 13, 2022 at 10:07 AM Brian Candler <[email protected]> wrote:
> Also, the whole expression can be simplified to:
>
> (node_memory_MemTotal_bytes{job="kubernetes-service-endpoints",cluster="$cluster",instance=~"$node"}
> - node_memory_MemFree_bytes) * on(instance)
> group_left(nodename) node_uname_info
>
> This is because you don't need to filter both node_memory_MemTotal_bytes
> and node_memory_MemFree_bytes. When you combine them with a binary
> operator (in this case subtraction), the result vector will only have
> values where both LHS and RHS have exactly matching labels. So any values
> from the RHS which don't match the LHS are discarded automatically, without
> having to explicitly filter them..
>
While that is semantically true, it can have performance implications
because the PromQL engine will first need to load *all* selected time
series (in this case, as many series as you have hosts) and only then throw
away the unmatched series. So if you have many hosts, this could get
inefficient.
> On Tuesday, 13 December 2022 at 09:03:36 UTC Brian Candler wrote:
>
>> If you want to do this inside prometheus, then you want recording rules
>> <https://prometheus.io/docs/prometheus/latest/configuration/recording_rules/>
>> .
>>
>> Note that with recording rules you're not really creating a new
>> "variable", you're creating a whole new timeseries. However this is what
>> you need, since your second query wants to perform aggregation_over_time()
>> on these precomputed values.
>>
>> It'll take a week until you have enough data from your recording rule to
>> use, unless you backfill
>> <https://prometheus.io/docs/prometheus/latest/storage/#backfilling-for-recording-rules>
>> .
>>
>> Your other option is to substitute the first expression into the two
>> places where you want to use it in the second expression. But in this
>> case, you'll need to use a subquery
>> <https://prometheus.io/docs/prometheus/latest/querying/basics/#subquery>,
>> e.g.
>>
>> (some - long - expression)[1w:1h]
>>
>> rather than
>>
>> metric[1w]
>>
>> The subquery shown evaluates (some - long - expression) over a 1w period
>> at 1h intervals to create a range vector. Note that unlike a simple metric
>> range vector, you need to give it the sampling interval. Or you can write
>>
>> (some - long - expression[1w:]
>>
>> but this will just use whatever global default evaluation interval you
>> have configured (which might be, say, 15 seconds or 1 minute)
>>
>> P.S. Are you sure that query you showed is valid? The first part is
>> fine, but the end doesn't make sense to me, and I can't see this syntax
>> permitted either by operators
>> <https://prometheus.io/docs/prometheus/latest/querying/operators/#group-modifiers>
>> or functions
>> <https://prometheus.io/docs/prometheus/latest/querying/functions/>.
>>
>>
>> node_memory_MemTotal_bytes{job="kubernetes-service-endpoints",cluster="$cluster",instance=~"$node"}
>> * on(instance) group_left(nodename)
>> *group(node_uname_info{})by(instance,nodename)*
>>
>> I would write this as simply:
>>
>>
>> node_memory_MemTotal_bytes{job="kubernetes-service-endpoints",cluster="$cluster",instance=~"$node"}
>> * on(instance) group_left(nodename) node_uname_info
>>
>> assuming that:
>> - N instances of the LHS match to 1 instance of the RHS with a
>> corresponding "instance" label
>> - you want to add the "nodename" label from the RHS to all the timeseries
>> returned by the LHS
>>
>> On Tuesday, 13 December 2022 at 08:03:32 UTC [email protected]
>> wrote:
>>
>>> Hello All,
>>>
>>> I wanted to store the below Prometheus query into an variable.
>>> (node_memory_MemTotal_bytes{job="kubernetes-service-endpoints",cluster="$cluster",instance=~"$node"}
>>> * on(instance) group_left(nodename)
>>> group(node_uname_info{})by(instance,nodename)-(node_memory_MemFree_bytes{job="kubernetes-service-endpoints",cluster="$cluster",instance=~"$node"}
>>> * on(instance) group_left(nodename)
>>> group(node_uname_info{})by(instance,nodename) ))
>>>
>>> and use it in second query ,
>>> z = (sum(rate(metric)[1m]) - sum(avg_over_time(metric[1w])) /
>>> sum(stddev_over_time(metric[1w]))
>>>
>>> Is there any way to achieve this.
>>> Thankyou
>>>
>>>
>>> --
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>
--
Julius Volz
PromLabs - promlabs.com
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